Question

Prove the Basic proportionality theorem.​

Answer 1

Basic Proportionality Theorem (can be abbreviated as BPT) states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. 



 

In the figure alongside, if we consider DE is parallel to BC, then according to the theorem,

ADBD=AECE

Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?

PROOF OF BPT

Given: In  ΔABC, DE is parallel to BC

Line DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.

To Prove: ADBD=AECE

Construction: Join segments DC and BE

Proof: 

In ΔADE and ΔBDE,

A(ΔADE)A(ΔBDE)=ADBD                 (triangles with equal heights)

In ΔADE and ΔCDE,

A(ΔADE)A(ΔCDE)=AECE                  (triangles with equal heights)

Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,

A(ΔBDE)=A(ΔCDE)

Therefore,

A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)

Therefore,

ADBD=AECE

Hence Proved.

Answer 2

Mark it as BRAINLIEST ✨